Transition metals

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  • Created by: Kayliss71
  • Created on: 22-05-18 13:33
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  • Transition metals
    • Properties of transition metals
      • They all have a high density
      • They all have high melting and high boiling points
      • Their ionic radii are more or less the same
      • Some can be used as catalyts
      • They have coloured ions
    • Complex ions
      • Complex ions are metal ions surrounded by ligands
        • A ligand is an atom, ion or molecule that donates a pair of electrons to a central transistion metal ion to form a coordinate bond
          • A ligand must have at least one pair of lone electrons
          • Monodentate ligands - can only form one coordinate bond E.g H2O,NH3,CL-
          • Multidentate ligands- can form more than one coordinate bond, e.g EDTA(4-) has 6 lone pairs
          • Bidentate ligands- are multidentate ligands that can form two coordinate bonds E.g ethane-1,2-diamine
      • The coordination number is the number of coordinate bonds that are formed with the central metal ion
      • 6 Coordinate bonds mean an octahedral shape
      • 4 Coordinate bonds mean a tetrahedral shape or a square planar shape
      • Some silver complexes have 2 coordinate bonds which forms a linear shape
      • They have an overall charge or total oxidation state.
        • The oxidation state of the metal ion = the total oxidation state - the sum of the oxidation states of the ligands
      • Haem in Haemoglobin contains a multidentate ligand
        • If carbon monoxide the Haem swaps its water ligand for a carbon monoxide ligand forming carboxyhaem-oglobin
          • This is bad because carbon monoxide is a strong ligand and doesn't readily exchange iwth oxygen meaning that haemoglobin cant transport oxygen
      • Complex ions can show optical isomerism
        • This happens with octahedral complexes with three bidentate ligands
      • Cis-Trans isomers can form in octahedral and square planar complexes
    • Formation of coloured ions
      • Ligands split the 3d sub-level into two energy levels
        • Electrons tend to occupy the lower orbitals. to jump up to the higher orbitals they need energy equal to the gap (delta E). they get this energy from visible light
        • The energy absorbed when electrons jump can be worked out using a formula: delta E = Plancks constant x frequency of light absorbed = (plancks constant x the speed of light) / wavelength of light absorbed
        • The amount of energy needed to make the electrons jump depends on the central metal ion and its oxidation state, the lignads and the coordination number as these affect the size of the energy gap (delta E)
      • When visible light hits a transition metal some frequencies are absorbed when electrons jump up to higher orbitals
      • The rest of the frequencies are transmitted or reflected
      • For example       hydrated     [Cu(H2O)6}2+ ions absorb light from the red end of the spectrum but  the complex reflects blue light
      • The colour of a complex can be altered by any of the factors that affect the size of delta E : Change in oxidation state, Change in co-ordination number and changes in ligands
    • Substitution reactions
      • If the ligands are similar size and the same charge, then the coordination number if the complex ion doesn't change and neither does the shape
        • [CO(H2O)6](2+) + 6NH3 --> [CO(NH3)6](2+) + 6H2O the complex goes from pink to straw coloured
        • In some cases the substitution is only partial: [Cu(H2O)6](2+) + 4NH3 --> [Cu(NH3)4(H2O)2](2+) + 4H2O. the complex goes from blue to deep blue
      • If ligands are different sizes there is a change of coordination number and a change of shape
        • [CO(H2O)6](2+) + 4Cl-   [CoCL4](2-) + 6H20. the complex goes from pink to blue and also changes from octahedral to tetrahedral
        • [Cu(H2O)6](2+) + 4Cl- [CuCl4](2-) + 6H2O. the complex has gone from pale blue to yellow and has changed from an octahedral shape to a tetrahedral shape
        • [Fe(H2O)6](3+) + 4Cl- [FeCl4](-) + 6H2O. the complex has stayed yellow but its shape has gone from octahedral to tetrahedral
      • Different ligands have different bond strengths, substitution can be easily reversed unless the new complex ion is much stronger than the old one
      • Multidentate ligands can form more stable complexes than mondentate ligands so the change can be harder to reverse
      • The chelate effect explains this - when monodentate ligands are substituted with bidentate or multidentate ligands the number of particles in the solution increases- the more particles the greater the entropy. Reactions that result in an increase in entropy are more likely to occur
    • Variable oxidation states
      • Transition metals can exist in variable oxidation states
      • When you switch between oxidation states its a redox reaction, for example vanadium can exists in four oxidation states in solution
        • oxidation state = +5, Formula of ion = VO2(+) colour of ion = yellow
        • Oxidation state= +4, Formula of ion = VO(2+), colour of ion = Blue
        • Oxidation state= +3, Formula of ion = V(3+), Colour of ion = Green
        • Oxidation state = +2, Formula of ion = V(2+), Colour of ion = violet
      • Vanadium (V) ions can be reduced by adding them to zinc metal in an acidic solution
        • 1) solution turns from yellow to blue as vanadium (V) is reduced to vanadium (IV)
          • 2VO2(+) + Zn + 4H(+) --> 2VO(2+) + Zn(2+) + 2H2O
        • 2) The solution then changes from blue to green as Vanadium(IV) is reduced to Vanadium (III)
          • 2VO(2+) + Zn + 4H(+) --> 2V(3+) + Zn(2+) + 2H2O
        • 3) The solution then changes from green to violet as Vanadium(III) is reduced to Vanadium (II)
          • 2V(3+) + Zn --> 2V(2+) + Zn(2+)
      • Redox potentials tell you how easy it is to reduce an ion. The larger the redox potential, the less stable the ion will be and so the more likely it is to be reduced
        • Redox potentials can vary depending on the environment that the ion is in.
          • Different ligands may make the redox potential larger or smaller depending on how well they bind to the metal ion in a particular oxidation state
          • Some ions need H(+) ions to be present in order to be reduced whilst others release (OH-) ions into solution when they are reduced. For reactions such as these the pH of the solution affects the size of the redox potential
            • In general redox potentials will be larger in more acidic conditions making the ion more easily reduced
      • Tollens reagent uses a reduction reaction to distinguish between aldehydes and ketones
        • RCHO + 2[Ag(NH3)2](+) + 3OH(-) --> RCOO(-) + 2Ag + 4NH3 + 2H2O
    • Catalysts
      • Transition metal catalysts work by changing oxidation states
      • In the Contact Process vanadium (V) oxide is able to oxidise SO2 to SO3 because it can be reduced to vanadium(IV) oxide
        • V2O5 + SO2 -> V2O4 + SO3. then the catalyst is oxidised back to its original state. V2O4 + 1/2 O2 --> V2O5
      • Heterogeneous catalyst are in different a phase from the reactants
        • The reaction takes place on active sites located on the surface of the catalyst. Support mediums are often used to make the area of the catalyst as large as possible , they help minimise cost as only a small coating of a catalkyst is needed to provide a large surface area
        • Iron acts as a catalyst in the Haber process for making ammonia. N2 + 3H2 --> 2NH3
        • Vanadium (V) oxide acts as a catalyst in the contact process for making sulfuric acid. SO2 + 1/2O2 -> SO3
        • Can be poisoned by impurities. these catalysts work by adsorbing reactants onto active sites located on their surfaces impurities may also bind and block the active sites and block reactants from being adsorbed
          • Catalyst poisoning reduces surface area which increases cost of the process as less product can be made in a certain time with a certain amount of energy
      • Homogeneous catallysts are in the same phase as the reactants.
        • They work by combining with the reactants to form an intermediate species which then reacts to form the products and re form the catalyst
      • Fe(2+) catalyses the reaction between S2O8(2-) and I(-)
        • Without a catalyst the ions repel each other so its unlikely they will collide and react
        • First the Fe(2+) ions are oxidised. S2O8(2-) + 2Fe(2+) --> 2Fe(3+) + 2SO4(2-)
        • Now the newly formed intermediate iron 3+ ions easily oxidise the I- ions to iodine and the catalyst is regenerated.  2Fe(3+) + 2I(-) --> I2 + 2Fe(2+)
      • Autocatalysis is when a product catalyses the reaction
        • 1) Mn(2+) catalyses the reaction by first reacting with MnO4(-)
          • MnO4(-) + 4Mn(2+) + 8H(+) --> 5Mn(3+) + 4H2O
        • 2) The newly formed Mn(3+) ions react with C2O4(2-) ions to form carbon dioxide and reform the catalyst ions
          • 2Mn(3+) + C2O4(2-) --> 2Mn(2+) + 2CO2
    • Metal aqua ions
      • Metal ions become hydrated in water as the water molecules form coordinate bonds with the metal ion
      • Solutions containing metal aqua ions are acidic. There is a reaction between the metal aqua ion and the water - this is a hydrolysis or acidity reaction
        • Cu(H2O)6(2+) + H2O [Cu(OH)(H2O)5](+) + H3O(+)
      • Metal 3+ ions form more acidic solutions because they have a high charge density and are much more polarising than the 2+ ions. more polarising power means that they attract electrons from the oxygen atoims of the coordinated water molecules more strongly, weakening the OH bond. so its more likely that a hydroigen ion will be released
      • Metal hydroxides can be amphoteric
        • They can act as both acids or bases. Aluminium hydroxide is amphoteric.  Al(OH)3(H2O)3
          • With acid:    [Al(H2O)6](3+) + 3H20
          • With base:   [Al(OH)4(H2O)2](-) + H2O
      • All four metal aqua ions will form precipitates with sodium hydroxide but only Aluminium hydroxide will dissolve because its amphoteric
        • All four metal ions will form precipitates with ammonia but only copper hydroxide precipitate will dissolve in an excess of ammonia (as it undergoes ligand exchange)
          • Cu(OH)2(H20)4 + 4NH3 [Cu(NH3)4(H2O)2](2+) + 2OH- + 2H2O
        • All four metal aqua ions will form precipitates with sodium carbonate but the solutions containing Al3+or Fe3+ will also form bubbles as CO2
          • [M(H2O)6](2+) + CO3(2-) MCO3 + 6H2O
            • CO3(2-) + 2H3O(+) CO2 + 3H2O

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